The neutralization reaction among all the reactions would be the one between an acid and a base to produce salt and water.
<h3>What is a neutralization reaction?</h3>
It is a reaction involving an acid and a base to produce salt and water.
From the list of reactions, the only reaction that involves acid and a base with salt and water being the products is the first reaction.
Thus, the neutralization reaction is represented by the equation:
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The enthalpy of the reaction can be obtained from the enthalpies of formation as -16.2 kJ/mol.
<h3>What is the enthalpy of reaction?</h3>
We know that any time that there is a chemical reaction, there is an interaction that place between the reactants and the products and as such we are going to get new substances and these are the substances that I have referred to here as the products of the reaction.
In this case, we are asked to obtain the enthalpy change of the reaction. This tells us the heat that could have been absorbed or evolved in the reaction. We have to at this point know that the enthalpy change of the reaction gotten from;
Sum of enthalpy of the products - Sum of enthalpy of the reactants
ΔH = [(-484.5)] - [(-393.5) + (-74.8)]
ΔH = (-484.5) + 468.3
ΔH = -16.2 kJ/mol
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Answer:
0.0585 M
Explanation:
- Pb(NO₃)₂ (aq) + 2NaCl (aq) → PbCl₂ (s) + 2NaNO₃ (aq)
First we <u>calculate the inital number of moles of each reagent</u>, using the <em>given volumes and concentrations</em>:
- 0.255 M Pb(NO₃)₂ * 52.1 mL = 13.3 mmol Pb(NO₃)₂
- 0.415 M NaCl * 38.5 mL = 16.0 mmol NaCl
Then we <u>calculate how many Pb(NO₃)₂ moles reacted with 16.0 mmoles of NaCl</u>, using the <em>stoichiometric coefficients of the reaction</em>:
- 16.0 mmol NaCl *
= 8.00 mmol Pb(NO₃)₂
Now we <u>calculate the remaining number of Pb(NO₃)₂ moles after the reaction</u>:
- 13.3 mmol - 8.00 mmol = 5.30 mmol Pb(NO₃)₂
Finally we <em>divide the number of moles by the final volume</em> to <u>calculate the concentration</u>:
- 5.30 mmol / (52.1 mL + 38.5 mL) = 0.0585 M
Answer:
Element Lithium
Explanation:
The element with the highest second ionization energy is lithium. It belongs to the alkaline metal group I.e group one metals
It has the highest second ionization energy because it is very difficult to remove the electron from the 1s orbital.
Its atomic number is 3. The electronic configuration is 1s2 2S1
